If there are 4 marbles left over each time, then we can forget about them for now.
So the question is, what is the smallest number than can be divided into 6,7 and 8?
the numbers have only one non-1 divisor in common: both 6 and 8 are divisible by 3.
so for our purposes we can "delete" one 2 and ask:
what is the smallest number than can be divided into 3,7 and 8 ?
There are no more divisors in common, so we just have to multiply them: 3*7*8=21*8=168
and the 4 marbles "extra"? We add them to this sum.
the the smallest possible number in the box is 168+4=172.
Since you know the triangles are congruent/equal, you know that:
m∠A and m∠X are congruent and have the same angle, and so does:
m∠B and m∠Y
m∠C and m∠Z
A triangle is 180°. (the 3 angles have to add up to 180) To find m∠B, you can do this:
m∠A + m∠B + m∠C = 180°
21° + m∠B + 35° = 180° Subtract 21 and 35 on both sides
m∠B = 124° your answer is C
(-3,2)
(-2, 1)
(-1, 0) <-----
(0, -1)
(1, -2)
Midpoint is (-1, 0)
Answer:
f(x)=x−2 f ( x ) = x - 2 , g(x)=x+2 g ( x ) = x + 2. Set up the composite result function. g(f(x)) g ( f ( x ) ). Evaluate g(f(x)) g ( f ( x ) ) by substituting in the value of f f ...
Step-by-step explanation:
